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Chapter 9: Problem 76

A thin, rigid, uniform rod has a mass of \(2.00 \mathrm{kg}\) and a length of \(2.00 \mathrm{m} .\) (a) Find the moment of inertia of the rod relative to an axis that is perpendicular to the rod at one end. (b) Suppose all the mass of the rod were located at a single point. Determine the perpendicular distance of this point from the axis in part (a), such that this point particle has the same moment of inertia as the rod does. This distance is called the radius of gyration of the rod.

Short Answer

Expert verified
(a) The moment of inertia is \(2.67 \mathrm{kg \cdot m^2}\); (b) the radius of gyration is approximately \(1.155 \mathrm{m}\).

Step by step solution

01

Understanding the Moment of Inertia

The moment of inertia for a uniform rod about an axis perpendicular to the rod through one end is given by the formula \( I = \frac{1}{3} mL^2 \). Here, \( m \) is the mass of the rod and \( L \) is its length. For this rod, \( m = 2.00 \mathrm{kg} \) and \( L = 2.00 \mathrm{m} \).
02

Calculating Moment of Inertia

Substitute the values of mass and length into the formula: \( I = \frac{1}{3} \times 2.00 \times (2.00)^2 \). Calculate the result: \( I = \frac{1}{3} \times 2.00 \times 4.00 = \frac{8}{3} \). Therefore, \( I = 2.67 \mathrm{kg \cdot m^2} \).
03

Concept of Radius of Gyration

The radius of gyration \( k \) is defined so that \( I = m k^2 \), where \( I \) is the moment of inertia and \( m \) is the mass of the object. We have \( I = 2.67 \mathrm{kg \cdot m^2} \) and \( m = 2.00 \mathrm{kg} \).
04

Solving for Radius of Gyration

Using the radius of gyration formula, substitute the known values for \( I \) and \( m \): \( 2.67 = 2.00 \cdot k^2 \). Solve for \( k^2 \): \( k^2 = \frac{2.67}{2.00} \). Calculate \( k \): \( k = \sqrt{1.335} \approx 1.155 \mathrm{m} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Uniform Rod
A uniform rod is a type of object that has consistent mass distribution along its length. This means that every segment of the rod has the same mass per unit length. Such rods are often used in physics problems because they simplify calculations associated with rotational dynamics, like the moment of inertia. When a rod is referred to as 'uniform', it implies that the rod is the same material throughout, providing a basis for straightforward mathematical modeling.
  • Each tiny segment of a uniform rod contributes equally to overall physical properties.
  • For practical purposes, this means that calculations involving the rod's rotation can assume uniform density.
Understanding how mass is evenly distributed helps accurately determine how the rod will behave when it's rotated around an axis.
Axis of Rotation
An axis of rotation is an imaginary line around which an object rotates. In the context of our problem, the axis is perpendicular to the rod and passes through one of its ends. This setup dictates how you calculate the moment of inertia, which is essential for predicting how the rod will rotate.
  • The placement of the axis affects the distribution of mass in relation to this point.
  • When the axis is at the end of the rod, the moment of inertia formula is specifically adjusted to factor in this particular geometry.
Choosing different axes can result in different values for the moment of inertia. Therefore, understanding the position of the axis is crucial for solving rotation-related physics problems accurately.
Radius of Gyration
The radius of gyration gives a useful measure that represents how the mass of an object is spread out relative to a particular axis of rotation. It is essentially the distance from the axis at which you could concentrate the mass into a point particle and still achieve the same moment of inertia.
  • The formula for finding radius of gyration is derived from moment of inertia relations, noted as: \( I = m k^2 \).
  • This formula helps redefine where the mass would be placed as if condensed to one point for simplicity.
While the radius of gyration itself doesn't affect the actual physical structure, it serves as a key computational tool in the analysis of rotating systems by simplifying complex mass distributions.

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Most popular questions from this chapter

A uniform board is leaning against a smooth vertical wall. The board is at an angle \(\theta\) above the horizontal ground. The coefficient of static friction between the ground and the lower end of the board is \(0.650 .\) Find the smallest value for the angle \(\theta\), such that the lower end of the board does not slide along the ground.

A solid cylindrical disk has a radius of \(0.15 \mathrm{m}\). It is mounted to an axle that is perpendicular to the circular end of the disk at its center. When a \(45-N\) force is applied tangentially to the disk, perpendicular to the radius, the disk acquires an angular acceleration of \(120 \mathrm{rad} / \mathrm{s}^{2} .\) What is the mass of the disk?

Two thin rods of length \(L\) are rotating with the same angular speed \(\omega\) (in \(\mathrm{rad} / \mathrm{s}\) ) about axes that pass perpendicularly through one end. Rod \(\mathrm{A}\) is massless but has a particle of mass \(0.66 \mathrm{kg}\) attached to its free end. Rod B has a mass of 0.66 kg, which is distributed uniformly along its length. The length of each rod is \(0.75 \mathrm{m},\) and the angular speed is \(4.2 \mathrm{rad} / \mathrm{s}\). Find the kinetic energies of rod \(A\) with its attached particle and of rod \(B\).

Two spheres are each rotating at an angular speed of \(24 \mathrm{rad} / \mathrm{s}\) about axes that pass through their centers. Each has a radius of \(0.20 \mathrm{m}\) and a mass of 1.5 kg. However, as the figure shows, one is solid and the other is a thin-walled spherical shell. Suddenly, a net external torque due to friction (magnitude \(=0.12 \mathrm{N} \cdot \mathrm{m}\) ) begins to act on each sphere and slows the motion down. Concepts: (i) Which sphere has the greater moment of inertia and why? (ii) Which sphere has the angular acceleration (a deceleration) with the smaller magnitude? (iii) Which sphere takes a longer time to come to a halt? Calculations: How long does it take each sphere to come to a halt?

A block (mass \(=2.0 \mathrm{kg}\) ) is hanging from a massless cord that is wrapped around a pulley (moment of inertia \(=1.1 \times 10^{-3} \mathrm{kg} \cdot \mathrm{m}^{2}\) ), as the drawing shows. Initially the pulley is prevented from rotating and the block is stationary. Then, the pulley is allowed to rotate as the block falls. The cord does not slip relative to the pulley as the block falls. Assume that the radius of the cord around the pulley remains constant at a value of \(0.040 \mathrm{m}\) during the block's descent. Find the angular acceleration of the pulley and the tension in the cord.
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